Re: uniqueness of fibonacci number extension

On 06.05.2008 15:30, David C. Ullrich wrote:
On Tue, 6 May 2008 02:00:02 +0000 (UTC), Jannick Asmus
<jannick.news@xxxxxx> wrote:

On 05.05.2008 21:00, G. A. Edgar wrote:
In article <fvmuka$h9$1@xxxxxxxxxxxxxxxxx>, Jannick Asmus
<jannick.news@xxxxxx> wrote:

On 05.05.2008 03:30, bo198214 wrote:
We know that there is a closed form for the Fibonacci function
(1) F(n)=F(n-1)+F(n-2), F(0)=0, F(1)=1
given by
F(n)=(r^n+(1-r)^n)/sqrt(5), where r=(1+sqrt(5))/2

Actually this is an extension from the domain of definition being the
natural numbers to real or even complex numbers. The extension is
analytic and satisfies equations (1) for all complex n.
Some correction is needed here: For the definition of r^z, z complex,
there is a (branch) of the logarithm needed which does not exist on the
complex plane (but on every simply connected open subset). So (1) cannot
hold if the is meant in terms of holomorphic functions.
You only need a log of r and of 1-r. So choose such logarithms, once
at the start, and use F(n)=(r^n+(1-r)^n)/sqrt(5) ... No worry about
branches, F(n) is defined for n in the whole complex plane.

Actually, F(n)=(r^n+(1-r)^n)/sqrt(5) isn't the Fibonaccis. Should
be F(n)=(r^n-(1-r)^n)/sqrt(5).

Different choices of these logs will give you different solutions,
answering negatively the question of uniqueness, right?
Thanks for making this a lot clearer. I believe such a function does not
even exist by the following reasoning with brute force:

The logarithms of r are ln(r)+2iPi.Z and of 1-r they are ln(r-1)+iPi+2iPi.Z.

Now let us assume that there is a choice of the logarithms of r and 1-r
such that (1) is satisfied for all complex figures, then in particular
for all positive reals x>2. Writing down relation (1) and sorting by
terms r^x, r^(x-1), r^(x-2) and (r-1)^x,... shows that this is
impossible unless exp(iPi.x) has real values only. Contradiction.

??? There is in fact no problem using (1) to define an entire
function. If you write down the details of whatever argument
you have in mind you'll find a flaw.

Right - G.A. provided a link to an explicit solution. And you will give
a family of those below.

Meanwhile I have thrown the piece of paper with my notes away, so that I
will not write it up again to find out where I went wrong. Dust bin was
just the right place for it anyway.

Given a complex number r <> 0 choose a complex number
L such that exp(L) = a. Now define p(z) = exp(Lz). This
is the function commonly denoted by r^z. There are infinitely
many such finctions, of course, one for each possible value
of L.

But regardless of the choice of L, it's easy to verify that
p(z+w) = p(z) p(w) for all complex z and w. And it
follows that p(n) = r^n for all integers n. And hence that
p(z+n) = p(z) r^n for all complex z and integral n.

And it follows that if F(z)=(r^z+(1-r)^z)/sqrt(5)
(or whatever the correct version is) then
F(z) = F(z-1) + F(z-2).

Right, too. I see now that the p's just separate leaving the
characteristic equation of r and 1-r.

My question:
Are there research results which give conditions that makes this
extension unique under all analytic extensions that also satisfy
equations (1) for all complex n? Or is this extension already unique
be these requirements?

For this question consider the vector space V over the complex numbers
with entire functions f(z) satisfying (1) and f(0)=f(1)=0. This implies
that f(Z)=0.

Then r^z.q(z)^n and (1-r)^z.q(z)^m with q(z)=exp(2iPi.z), m,n integers,
are in V and linearly independent. Hence there is no uniqueness.

But this was already pointed out by David basically, including the
reference to Carlson's theorem on uniqueness if f is of exponential type.

--
Best wishes,
J.

.

Relevant Pages

  • Re: VOTE on whether 1/oo = 0
    ... the poll chose to answer using only a context in which 1/oo is nonsensical, ... But the two-point extension of the reals is ... need not be regarded as being merely an abbreviation. ...
    (comp.theory)
  • Re: VOTE on whether 1/oo = 0
    ... the poll chose to answer using only a context in which 1/oo is nonsensical, ... But the two-point extension of the reals is ... need not be regarded as being merely an abbreviation. ...
    (sci.math)
  • Re: VOTE on whether 1/oo = 0
    ... the poll chose to answer using only a context in which 1/oo is nonsensical, ... But the two-point extension of the reals is ... need not be regarded as being merely an abbreviation. ...
    (sci.logic)
  • Re: Point at infinity in the extended complex number system
    ... The part about endless intervals seems confused. ... arithmetic in the two-point extension of the reals is quite surprising. ... arithmetics in the two systems. ...
    (sci.math)
  • Re: Determining MS Office docs
    ... Hi David, ... Prasad ... > and only one application to a given extension. ... > user to generate multiple associations? ...
    (comp.programming)